# How do you simplify (c^3d^4)/(cd^7)?

Aug 31, 2017

See a solution process below:

#### Explanation:

First, rewrite the expression as:

$\left({c}^{3} / c\right) \left({d}^{4} / {d}^{7}\right)$

Next, use these rules of exponents to simplify the $c$ terms:

$a = {a}^{\textcolor{b l u e}{1}}$ and ${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} - \textcolor{b l u e}{b}}$

$\left({c}^{3} / {c}^{\textcolor{b l u e}{1}}\right) \left({d}^{4} / {d}^{7}\right) \implies \left({c}^{\textcolor{red}{3}} / {c}^{\textcolor{b l u e}{1}}\right) \left({d}^{4} / {d}^{7}\right) \implies {c}^{\textcolor{red}{3} - \textcolor{b l u e}{1}} \left({d}^{4} / {d}^{7}\right) \implies {c}^{2} \left({d}^{4} / {d}^{7}\right)$

Now, use this rule of exponents to simplify the $d$ terms:

${x}^{\textcolor{red}{a}} / {x}^{\textcolor{b l u e}{b}} = \frac{1}{x} ^ \left(\textcolor{b l u e}{b} - \textcolor{red}{a}\right)$

${c}^{2} \left({d}^{\textcolor{red}{4}} / {d}^{\textcolor{b l u e}{7}}\right) \implies {c}^{2} \left(\frac{1}{d} ^ \left(\textcolor{b l u e}{7} - \textcolor{red}{4}\right)\right) \implies {c}^{2} \cdot \frac{1}{d} ^ 3 = {c}^{2} / {d}^{3}$

Aug 31, 2017

$\frac{{c}^{2}}{{d}^{3}}$

#### Explanation:

$\text{using the "color(blue)"law of exponents}$

•color(white)(x)a^m/a^n=a^(m-n)to"for "m>n

•color(white)(x)a^m/a^n=1/a^(n-m)to" for "n>m

$\frac{{c}^{3} {d}^{4}}{c {d}^{7}}$

$= {c}^{3} / {c}^{1} \times {d}^{4} / {d}^{7}$

$= {c}^{\left(3 - 1\right)} \times \frac{1}{d} ^ \left(\left(7 - 4\right)\right)$

$= {c}^{2} / 1 \times \frac{1}{d} ^ 3 = {c}^{2} / {d}^{3}$